The use of the dx in the quantum state vector integral

[etotheipi]

As a simple example, the probability of measuring the position between $x$ and $x + dx$ is $|\psi(x)|^{2} dx$ since $|\psi(x)|^{2}$ is the probability density. So summing $|\psi(x)|^{2} dx$ between any two points within the boundaries yields the required probability.

The integral I'm confused about is the expansion of the quantum state vector, like so$$\left| \psi \right> = \int \psi(x) \left| x \right> dx$$If the wavefunction were defined between 0 and 1, and I were to take the left Riemann sum (not rigorously, but just vaguely for explanation purposes), I would get $$\left| \psi \right> = \psi(0) \left| 0 \right> dx + \psi(dx) \left| dx \right> dx + \psi(2dx) \left| 2dx \right> dx + ...$$What is the purpose of the trailing $dx$ in each term (i.e. which part does it multiply to), or is this the completely wrong way to think about it?

 

[vanhees71]

This makes use of the socalled completeness relation,
$$\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=\hat{1},$$
i.e., you can decompose any ket $|\psi \rangle$ in terms of the position-wave function, $\psi(x)=\langle x|\psi \rangle$,
$$|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x |x \rangle \psi(x).$$
It's like the decomposition of vectors in a finite-dimensional Euclidean vector space with respect to some Cartesian basis $\vec{e}_j$. A vector $\vec{V}$ has components $V_j=\vec{e}_j \cdot \vec{V}$, and you can decompose the vector in terms of this basis again as
$$\vec{V}=\sum_{j} (\vec{e}_j \cdot \vec{V}) \vec{e}_j = \sum_{j} V_j \vec{e}_j.$$
Also here you have a completeness relation in terms of the tensor product
$$\sum_j \vec{e}_j \otimes \vec{e}_j=\hat{1}.$$